Question 14

Identify X: $CH_3I \xrightarrow{KCN} CH_3CN \xrightarrow{H_3O^+} CH_3COOH \xrightarrow{P, Br_2} BrCH_2COOH \xrightarrow{NH_3} NH_2CH_2COOH$ (Glycine).

Molar mass of Glycine ($C_2H_5NO_2$) = $2(12) + 5(1) + 14 + 32 = 75$ g/mol.

Identify Y: Hydrolysis of caprolactam yields $\epsilon$-aminocaproic acid ($NH_2(CH_2)_5COOH$).

Molar mass ($C_6H_{13}NO_2$) = $6(12) + 13(1) + 14 + 32 = 131$ g/mol.

Formation of Z: $500$ mol X + $500$ mol Y reacts to form $1$ mol of copolymer Z (Nylon-2-nylon-6). Total monomers = $500 + 500 = 1000$. Number of peptide bonds in a single acyclic chain of $1000$ units = $1000 - 1 = 999$.

Each bond formation releases one $H_2O$ molecule ($18$ g/mol).

Calculation: Mass of $1$ mol Z = (Total mass of reactants) - (Total mass of water lost)

Mass = $(500 \times 75) + (500 \times 131) - (999 \times 18)$

Mass = $37500 + 65500 - 17982 = 103000 - 17982 = 85018$ g.